If `(sin^(-1)x)^2+(cos^(-1)x)^2=(17pi^2)/(36)`, find `xdot`

1.	If `(sin^(-1)x)^2+(cos^(-1)x)^2=(17pi^2)/(36)`, find `xdot`
Answer» `(17 pi^2)/36 = (sin^-1 x + cos^-1 x)^2 - 2 sin^-1 x cos^-1 x` `(17 pi^2)/36 = pi^2/4 - 2sin^-1 x[ pi/2 - sin^-1 x]` `2(sin^-1 x)^2 - pi(sin^-1 x) + pi^2/4- 17 pi^2/36= 0` `2(sin^-1 x)^2 - pi(sin^-1 x) - (8pi^2)/36 = 0` `(sin^-1 x)^2 - pi/2 (sin^-1 x) - pi^2 / 9= 0` `sin^-1 x = (pi/2 +- sqrt(pi^2/4 + (4 pi^2)/9))/(2(1))` `sin^-1 x = (pi/2 +- sqrt((25 pi^2)/(36)))/2` now, `(pi/2 + 5pi/6)/2 or (pi/2 - 5 pi/6)/2` `8pi/24 = pi/3or -pi/3` `sin^-1 x = pi/3` `x = sqrt3/2` and `sin^-1 x = -pi/6` `x= -1/2` Answer

Discussion

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